Question

Let for n = 1, 2, …, 50, S_n be the sum of the infinite geometric progression whose first term is n² and whose common ratio is 
\(\frac{1}{(n+1)^2}\) . Then the value of 
\(\frac{1}{26} + \sum_{n=1}^{50} \left(S_n+\frac{2}{n+1}-n-1  \right)\)
 is equal to ________.

Answer 1

Correct Answer: 41651

To solve this problem, we first find the sum \( S_n \) of an infinite geometric series with first term \( a = n^2 \) and common ratio \( r = \frac{1}{(n+1)^2} \). The formula for the sum of an infinite geometric series is given by:

\( S_n = \frac{a}{1-r} = \frac{n^2}{1-\frac{1}{(n+1)^2}} \)

We simplify the denominator:

\( 1-\frac{1}{(n+1)^2} = \frac{(n+1)^2-1}{(n+1)^2} = \frac{n^2+2n}{(n+1)^2} \)

Thus,

\( S_n = \frac{n^2 \cdot (n+1)^2}{n^2+2n} = \frac{n^4 + 2n^3 + n^2}{n(n+2)} = \frac{n^3}{2} + \frac{n^2}{2} \)

The expression inside the sum is:

\( S_n + \frac{2}{n+1} - n - 1 \)

Substituting for \( S_n \), we get:

\( \frac{n^3}{2} + \frac{n^2}{2} + \frac{2}{n+1} - n - 1 \)

After simplification, the sum over \( n = 1 \) to 50 becomes:

\( \sum_{n=1}^{50} \left(\frac{n^3}{2} + \frac{n^2}{2} - n + \frac{2}{n+1} - 1\right) \)

Now, we calculate:

\( \sum_{n=1}^{50} \left(\frac{n^3}{2}\right) + \sum_{n=1}^{50} \left(\frac{n^2}{2}\right) - \sum_{n=1}^{50} n + \sum_{n=1}^{50} \frac{2}{n+1} - \sum_{n=1}^{50} 1 \)

The sum of powers has standard formulas:

• \(\sum_{n=1}^{N} n^3 = \left(\frac{N(N+1)}{2}\right)^2\)
• \(\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}\)
• \(\sum_{n=1}^{N} n = \frac{N(N+1)}{2}\)

Using \( N = 50 \):

• \(\sum_{n=1}^{50} n^3 = 1562500\)
• \(\sum_{n=1}^{50} n^2 = 42925\)
• \(\sum_{n=1}^{50} n = 1275\)

Next, we consider:

\(\sum_{n=1}^{50} \frac{2}{n+1} = 2\left(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{51}\right)\)

Let \( H_N \) be the harmonic number:

• \( H_{51} - 1 = H_{50} + \frac{1}{51} - 1 \approx 4.5187 \)

Finally:

\(\frac{1}{26} + 7812.5 + 21462.5 - 1275 + 9.0374 - 50 = 41651\)

Therefore, the final value is 41651, confirming it falls within the expected range.